Question

请问各位大神 这题怎么做啊 在线等 急急急！！！！！

Answer 1

∫(e->e^2) ln(r^2) r dr

=(1/2)∫(e->e^2) ln(r^2) dr^2
=(1/2)[r^2.ln(r^2)]|(e->e^2) -(1/2)∫(e->e^2) r^2.[1/(r^2)](2r) dr
=2e^4-e^2 - ∫(e->e^2) r dr
=2e^4-e^2 - (1/2)[r^2]|e->e^2)
=2e^4-e^2 - (1/2)(e^4-e^2)
=(3/2)e^4 -(1/2)e^2
∫(0->2π) dθ∫(e->e^2) ln(r^2) r dr
=∫(0->2π) [ (3/2)e^4 -(1/2)e^2 ] dθ
=( 3e^4 -e^2 ).π

追问

好厉害啊 这个题怎么写啊 救命

追答

∫(1->2) dx ∫(x->x^2) f(x,y) dy

=∫(1->4) dy ∫(√y->y) f(x,y) dx

D:

x≤y≤x^2

1≤x≤2

D:

1≤y≤4

√y≤x≤y

Answer 2

I = ∫<0, 2π>dθ∫<e, e^2>ln(r^2)rdr = 2π∫<e, e^2>2ln(r) rdr
= 2π∫<e, e^2>lnrd(r^2) = 2π[r^2lnr]<e, e^2> - 2π∫<e, e^2>r^2dlnr
= 2π(2e^2-e) - 2π∫<e, e^2>rdr = 2π(2e^2-e) - π[r^2]<e, e^2>
= 2π(2e^2-e) - π[e^2-e] = πe(3e-1)